CHAPTER 8 SCREWS, FASTENERS, NONPERMANENT JOINTS This chapter deals with the design and analysis of nonpermanent fasteners such as bolts, power screws, cap screws, setscrews, eys and pins. 8- Standards and Definitions A screw thread used for fastening and related terminology are shown below in the figure: Major Diameter, d Minor Diameter, d r or d Root Crest Pitch p Another important dimension not shown in the figure is the lead l, which is the distance that the nut advances along the screw axis when it is given one complete turn. We have l np, where n is the number of threads for multiple-threaded screws. Standard screws are single-threaded, and hence for them l p. There are two standards for screw threads, which are Unified (UN, Table 8-, pg. 399) for inchtypes and Metric (M, Table 8-, pg. 398) for millimeter-types. In each type, there are Course (C) and Fine (F) threads. Furthermore, the MJ threads for the Metric and UNR threads for the Unified have rounded roots to improve their fatigue strengths. As an example, a M.75 thread would mean a Metric thread with a major diameter of d mm and a pitch of p.75 mm. Also, a 8 5 in 8 UNRF would indicate a Unified Fine thread with a rounded Root having a major diameter of d 8 5 in 0.65in and 8 threads in one inch of axial length. 8- Power Screws The power screws are used for power transmission, and although the basic terms are same as those of the fastening screws, nevertheless they have different threads. There are two types of threads: Square and ACME, which are shown below:
p/ p/ p/ p/ p/ Thread Angle d d α d r Square Thread d r ACME Thread Important considerations for the power screws are the torques needed for raising a load (T R ) and lowering a load (T L ), which are calculated for a square thread as F dm π f d + m l TR π dm f l and F dm π f d m l T L π dm + f l where d m is the mean diameter of the power screw given for a square thread as d m d p/. Also, F is the load to be raised or lowered and f is the coefficient of friction between the screw and nut. A power screw is said to be self-locing if the nut does not roll bac by itself after the load is raised. For the self-locing of a screw, we should have T L > 0, or, from the above equation π f d m > l. We also define an efficiency e for raising the load e T0 TR where T 0 T 0 TR f 0 is the torque for raising the load without the friction, i.e. F l, and hence π e F l π T R. The torques for raising and lowering the load for an ACME thread are found by multiplying the friction terms by α cosα sec, and thus F dm π f d + m secα l TR and π dm f l secα F dm π f d m secα l T L. π dm + f l secα For both the square and ACME threads, if there is a collar attached to the screw for the load to sit on, then the friction on this collar should also be included for finding
3 the torques of raising and lowering the load. We now have the total torque T for raising the load T T R + T c T R + Ff c d c where f c is the coefficient of friction between the collar and the load, and d c is the diameter of friction circle. For lowering the load, the total torque T is given by T T L + T c. The stresses on a power screw during raising or lowering a load are 3-dimensional. These stresses are given for a square thread as: d m p/ F p/ z x 4F ) Axial compressive stress: z, π dr 6T ) Torsional shear stress: τ yz, and 3 π dr 6F 3) Bending stress: x. This bending stress can also be approximated as π drnt p 6(0.38F) x, which assumes that only the st tooth between the nut and screw π dr p taes the 38% of the load. Hence, we can compute an equivalent von Mises stress from these 3 stresses using the equation (6-4) in Chapter 6, pg. 6, as d r [ + + ( ) 6τ ] / x z z x + This equivalent stress can be used to find a factor of safety for the power screw to decide on its strength. There is another stress, bearing stress B, that is used to judge if the force F is crushing the surface of the thread. This stress is calculated as yz
4 Note: Review Example 8-, pgs. 405-407. B F F π d n p / π d n p. m t 8-3 Standards and Definitions Various hexagonal-head bolts and cap screws are shown in Figures 8-9 and 8-0, pg. 409, in the textboo. Several machine screw head types are illustrated in Figure 8-, pg. 40. Dimensions of hexagonal bolts and cap screws are given in the appendix of the textboo, Tables A-9 and A-30, pgs. 007 and 008. Hexagonal nuts are shown in Figure 8-, pg. 40, and their dimensions are provided in the appendix, Table A- 3, pg. 009. 8-4 Modeling of Screw Joints Fastener (Bolt or Screw) Stiffness A screw joint or connection is idealized through parallel springs with a bolt stiffness (spring constant) of b and member stiffness of m as shown in the figure below: m t b m The bolt stiffness b is found using the length of the screw within the grip of the connection. The model of the bolt is thought as two springs in series, one spring corresponding to the unthreaded portion and the other spring for the threaded portion of the bolt. So, we have b d + t where d and t are the stiffnesses of the unthreaded and threaded portions of the screw, which are defined as Ad E E d and t ld lt where π A d cross-sectional area of unthreaded portion 4 d (d is major diameter), l d length of unthreaded portion, A t tensile stress area found in Tables 8- and 8- (pgs. 398 and 399), and l t length of threaded portion of the screw within the grip. Hence, we have from the above equations: Ad E b. Ad lt + ld
5 8-5 Modeling of Screw Joints Member Stiffness The calculation of the stiffness m for the members clamped by a bolt is not as straightforward. Members are clamped in series, hence we have m + +K The stiffness for each member in the above equation (,, and so on) can be found from 0. 5774π E d (. 55t + D d)( D + d) ln (. 55t + D + d)( D d) where t is the thicness of the member, d is the major bolt diameter and D.5d. If there is a soft gaset among the members, then we can set as m g, where g is the gaset stiffness. If there are only two clamped members that are identical, i.e. they have the same thicness of t and are made up of the same material, then the resultant member stiffness is obtained directly from m 0. 5774π E d (0. 5774l + 0. 5d) ln 5 (0. 5774l +. 5d) where lt. In the case of two identical members we can also use the equation m Bd / l Ae Ed where the constants A and B can be obtained from Table 8-8, pg. 46, in the textboo. Note: Review Example 8-, pgs. 46 and 47. 8-6 Bolt Strength Material Properties for Bolts A bolt strength is specified by its proof and tensile strengths, which are listed in Tables 8-9, 8-0 (UN bolts) and 8- (M bolts), pgs. 48 through 40 in the textboo. A proof load is the maximum load or force that a bolt can withstand without becoming plastic. Then the proof strength is defined as the proof load divided by the bolt s tensile stress area. 8-7 and 8-9 Static Analysis of Bolts The scenario is as follows. We first apply a preload of F i to the bolt, which subjects the bolt to tension and clamped members to compression. Afterwards, the joint is put to service, where both the bolt and members are subjected to an external load of P. This P is taen by the bolt as P b and by the members as P m. Remember from the previous Section 8-4 that the bolt and members are connected as two parallel springs of b and m. We can find the following relations
6 b m Pb P C P and Pm P ( C) P b + m b + m where C is called the joint constant. Consequently we have and F b resultant bolt load P b + F i CP + F i F m resultant member load P m F i ( C)P F i The members should always be compressed and hence we always need F m < 0. The preload F i in the above equations is assumed as 0. 75 S Fi 0. 90 S p p for for reused bolts or screws permanent bolts or screws Now we can calculate the bolt stress as Fb b C P + Fi Multiplying the external load P by a load factor of n and letting the bolt stress equal to its proof strength at the limit, we get C np + which we can solve it for the load factor as Fi S p S p Fi n. C P Remember that always n > and it stands for a ind of factor of safety for the bolt strength. We should also chec the member tightness with another factor of safety n 0 to guard against the joint separation. We have for the members at the limit of separation from which we obtain ( C) n 0 P F i 0 And again we need n 0 >. n 0 Fi ( C) P. Note: Review Example 8-4, pgs. 47 and 48.
7 8-8 Required Bolt Torque We can express the torque T required to produce a preload F i by T KF i d where K is obtained for different bolt conditions from Table 8-5, pg. 44. In the absence of any information we can assume that K 0.. Note: Review Example 8-3 (parts a and b), pgs. 44 and 45. 8- Fatigue Analysis For the fatigue, we assume that the external load changes between 0 and P. Hence we have the maximum and minimum bolt forces as (F b ) max CP + F i and (F b ) min F i Then the maximum and minimum bolt stresses are found as ( b ) max C P Fi C P + i A t + and ( b ) min F i i Thus, we have the alternating and midrange stresses in the bolt as and m a ( b ) max ( b ) min ( b ) max + ( b ) min C P + C P A t i a + i Now we can use one of the fatigue methods to find the fatigue factor of safety n f for the bolt. If we use the Gerber approach n f a Se n f m + Sut which can be solved for n f. The endurance limit or endurance strength S e in the above equation for the bolt can be obtained directly from Table 8-7, pg. 430, in the textboo. Remember that we get the tensile strength S ut for the bolt from Tables 8-9 through 8-, pgs. 48-40. Or, we can find the fatigue factor of safety from n f S a / a, where S a is given as S + + a Sut Sut 4Se( Se i ) Sut ise Se. Note: Review Example 8-5 (only concern about n, n 0 and n f ), pgs. 43-435.
8 8- Shear Joints When a group of bolts or pins are subjected to shear, we have to find the resultant shear forces and stresses on the bolts and identify the critical bolt or bolts taing the maximum shear stress. A typical situation is shown below: l F l Beam F A G A F B A F C F C F A r C F C C r A G There are four bolts or pins at the left side of the beam, which are resisting the reaction force of V and the moment of M Fl. Reactions V and M are put in the centroid of the left bolt group, G. We have two types of shear forces in each bolt, one is the primary shear force ( F A, F B, F C and F D ) and the other is the secondary shear force ( F A, F B, F C and F D ). The primary shear forces are due to the shear force of V and hence are equal to V r B F D F D F B M rd D F D F B F B F A F B F C F D V n where n is the number bolts in the bolt group (it is 4 in this case). The secondary shear forces exist to resist the bending moment M and thus we have F A M ra ra + rb + rc + rd and so on. As shown in the figure, we have to find out the resultant shear force in each bolt (F A, F B, F C, and F D ) and divide it by the cross-sectional area to calculate the shear stress. It loos lie in the figure that the bolts B and D are taing the maximum shear forces. Hence, assuming that all the bolts are of the same size (diameter), the bolts B and D are the critical ones, because they get the maximum shear stresses. Note: Review Example 8-7, pgs. 439 and 440.
9 8-3, 4 Setscrews, Keys and Pins These machine elements are used to mount a hub (dis, gear, pulley, etc) on a shaft. Setscrews rely on compression to develop a clamping force between the hub and the shaft. This clamping force generates a friction force along the contact surface, which holds the hub axially on the shaft. This is in a way similar to a press and shrin fit. Several types of setscrews are shown in Figure 8-7 and setscrew sizes are given in Table 8-8, pg. 44, in the textboo. Keys and pins are also used for mounting hubs onto shafts (see Figure 8-8, pg. 44). Keys are categorized according to their cross-sections: rectangular, square and round eys. There are also special eys: Gib-head and Woodruff eys. The eys are placed inside grooves between the shaft and hub to provide axial positioning of the hub on the shaft, whereas the pins are inserted inside transverse holes in shafts and hubs. The pins are also classified with respect to their shapes: round and taper pins. Some ey and pin dimensions are listed in Tables 8-9 and 8-0, pgs. 44 and 443. Note: Review Example 8-8, pg. 446.